/*
The size of the hash table is not determinate at the very beginning. If the total size of keys is too large (e.g. size >= capacity / 10), we should double the size of the hash table and rehash every keys. Say you have a hash table looks like below:

size=3, capacity=4

[null, 21, 14, null]
       ↓    ↓
       9   null
       ↓
      null
The hash function is:

int hashcode(int key, int capacity) {
    return key % capacity;
}
here we have three numbers, 9, 14 and 21, where 21 and 9 share the same position as they all have the same hashcode 1 (21 % 4 = 9 % 4 = 1). We store them in the hash table by linked list.

rehashing this hash table, double the capacity, you will get:

size=3, capacity=8

index:   0    1    2    3     4    5    6   7
hash : [null, 9, null, null, null, 21, 14, null]
Given the original hash table, return the new hash table after rehashing .

Example
Given [null, 21->9->null, 14->null, null],

return [null, 9->null, null, null, null, 21->null, 14->null, null]

Note
For negative integer in hash table, the position can be calculated as follow:

C++/Java: if you directly calculate -4 % 3 you will get -1. You can use function: a % b = (a % b + b) % b to make it is a non negative integer.
Python: you can directly use -1 % 3, you will get 2 automatically.
Tags Expand 
LintCode Copyright Hash Table

Thoughts:
1. Loop through the hashtable[] and find longest, calcualte new capacity
2. rehash

*/

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
 public class Solution {
    /**
     * @param hashTable: A list of The first node of linked list
     * @return: A list of The first node of linked list which have twice size
     */    
    public ListNode[] rehashing(ListNode[] hashTable) {
    	if (hashTable == null || hashTable.length == 0) {
    		return hashTable;
    	}
    	//Find longest size
    	/*
    	int longest = 0;
    	for (int i = 0; i < hashTable.length; i++) {
    		ListNode node = hashTable[i];
    		int count = 0;
    		while (node != null) {
    			count++;
    			node = node.next;
    		}
    		longest = Math.max(longest, count);
    	}*/
    	//Calculate new capacity
    	//Just to clarify, this problem asks to double the hashtable size, rather than 'longest' times longer.
    	int capacity = hashTable.length * 2;
    	if (capacity == hashTable.length) {
    		return hashTable;
    	}
    	
    	ListNode[] rst = new ListNode[capacity];
    	for (int i = 0; i < hashTable.length; i++) {
    		ListNode node = hashTable[i];
    		while (node != null) {
    			ListNode newNode = new ListNode(node.val);
				int hCode = hashcode(newNode.val, capacity);
				if (rst[hCode] == null) {
    				rst[hCode] = newNode;
    			} else {
    				ListNode move = rst[hCode];
    				while (move.next != null) {
    					move = move.next;
    				}
    				move.next = newNode;
    			}
    			node = node.next;
    		}
    	}

    	return rst;
    }

    public int hashcode(int key, int capacity) {
    	if (key < 0) {
    		return (key % capacity + capacity) % capacity;
    	} else {
    		return key % capacity;
    	}
    }
};













